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查看: 2099|回复: 12
 
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[問]有關logic circuit的問題!!!
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問題是用英文寫的...可以話,請幫忙翻譯翻譯...  
 
Design a 4bit synchronous counter with a count sequence based on a number "6525".This number is our count sequence. 
 
*注* Use a minimum number of J-K type flip-flops and NAND Gates in the design.Any appropriate logic minimisation technique can beused. Note that flip-flops have Q and Qbar outputs. |   
 
 
 
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发表于 2-6-2004 11:48 AM
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Pop^o网友, 
找到一个不错的网站: 
http://www.play-hookey.com/digital/ripple_counter.html 
可看看counter的部分:basic 4-bit counter, synchronous binary counter, frequency  
divider等.. 
 
有点印象,不知有没有记错: 
(都还给老师了,现在也没有书参考) 
1.先算要用几个4-bit counter 
2. 用logic circuit,到了count 的数,要reset所有JK flip flop.  
3. logic circuit部分,minimize gates的数目..(很懒,没去想,哈哈). 
 
你作到那里了? 
以前念书时好像用一本O'Malley的书,好像有讲到.. |   
 
 
 
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发表于 3-6-2004 10:45 PM
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Synchronous circuit is driven by clock, and the flip flop will change state simultaneously in each clock cycle.4 bit meaning 4 jk ff, and you were asked to created the count sequence of 
 
A0101 
B1010 
C1101 
D0000 
 
with 1 represent rising edge and 0 falling edge. Since the most significant bit is low all the time, you actually need only 3 jk ff. (8 is easier than 16) 
 
The no. of Nand gate will be determined by your design technique ( k map etc). 
 
You can start with a state diagram, followed by state assignment, state table, transistion table and  excitation table. |   
 
 
 
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 楼主 |
发表于 3-6-2004 11:17 PM
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 clearskies 于 3-6-2004 22:45  说 :  
Synchronous circuit is driven by clock, and the flip flop will change state simultaneously in each clock cycle.4 bit meaning 4 jk ff, and you were asked to created the count sequence of 
 
A0101 
B ...   
Thanks man....but how about the self restarting problem? for eg if the first starting number isnot in my conting sequence, how do i overcome the problem? 
make assumption that all the non-counting sequnce number will go into my counting sequence? 
or...? |   
 
 
 
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发表于 4-6-2004 12:28 AM
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哈哈,不好意思,原来我误解了Pop^o网友的题目,是"6-5-2-5"的sequence,不是6525个sequence.. 
 
看来我真的忘得7788了.. 
 
B=NOT(A),可以用第一个FF的Q和Qbar,用2个FF有没有问题? 
if Q(FF1)=0101, Q(FF2)=0011, 
C=Q(FF1) OR Q'(FF2)= (0101)OR(1100)=1101 
这样可以吗? 
 
麻烦网友教教我general method 里的transistion table and excitation table那 
部分,我全忘了,又找不到书.. 
A        B             C       D 
011 ->101 ->010 ->101 
 
101又-->去011.. 
之后怎么作? |   
 
 
 
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发表于 4-6-2004 07:05 AM
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is it using the 7 LED decoder? i have forgotten the code. |   
 
 
 
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 楼主 |
发表于 4-6-2004 06:54 PM
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 calvin_tan 于 4-6-2004 07:05  说 :  
is it using the 7 LED decoder? i have forgotten the code.   
 
 
nono..... 
dats is call BCD....this onw is counter with JK ff.... 
is different |   
 
 
 
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发表于 4-6-2004 11:41 PM
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记忆一点点恢复了.. 
好像是这样: 
 
inputs to logic circuit 先假设用3个FF吧) 
Q(FF1)= 01010101 
Q(FF2)= 00110011 
Q(FF3)= 00001111 
Outputs: 
A=          01010101 
B=          10101010 
C=          11011101 
D=          00000000 
 
应该写成truth table的,这里从简写"横"的.. 
用Karnaugh map,可找出: 
A=Q(FF1) 
B=Qbar(FF1) 
C=Q(FF1) OR Qbar(FF2) 
所以不需FF3 
 
好像不会有self restarting problem.. |   
 
 
 
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 楼主 |
发表于 5-6-2004 02:39 AM
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 flyingfish 于 4-6-2004 23:41  说 :  
记忆一点点恢复了.. 
好像是这样: 
 
inputs to logic circuit 先假设用3个FF吧) 
Q(FF1)= 01010101 
Q(FF2)= 00110011 
Q(FF3)= 00001111 
Outputs: 
A=          01010101 
B=          10101010 
C=          ...   
 
 
不是很明白Q(FF1)= 01010101 
          Q(FF2)= 00110011 
          Q(FF3)= 00001111 
怎樣來的??? |   
 
 
 
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发表于 6-6-2004 12:34 PM
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 楼主 |
发表于 7-6-2004 06:37 PM
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发表于 12-6-2004 09:52 PM
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use moore machine or mealee machine will save u a lot of time |   
 
 
 
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 楼主 |
发表于 16-6-2004 12:47 AM
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